--- In

primenumbers@yahoogroups.com, "Mark Underwood"

<mark.underwood@s...> wrote:

> which would not make the devil happy ;) Incredibly your expression

> never yields prime factors below 59.

A quadratic expression that yields no prime factors less than a given

value is easy to generate arbitrarily.

Start with any integer valued f(x)=ax^2+bx+c and divide through by any

common factor. Let's say f(d) is odd and f(d+1) is even, then take

the three seed values f(d), f(d+2),f(d+4) and solve for the

coefficients of the new quadratic, f'(x) (you have 3 eqs, 3 unknowns,

so this can always be done).

If 3 appears as a factor of f'(x), only 2 of 3 consecutive solutions

of f'(x) can possibly be divisible by 3. Choose one that is not, say

it is f'(e), then take 3 consecutive values f'(e),f'(e+3),f'(e+6) and

solve for the quadratic f''(x). Continue as high as you please. Any

prime factor q, will appear at most twice within q consecutive values

of the quadratic progression, so there are always at least q-2

"paralell" quadratic progressions where q cannot be a divisor of the

quadratic.

-Dick Boland